Denote \(\mathpzc {p}_i=\mathpzc {p}_i(\lambda )\)  for brevity. The number of edges in a random graph \(G_n\)  of order \(n\)  is a binomial random variable with parameters \(N={n\atopwithdelims ()2}\)  and \(p\) . The expected number of edges in \(G_n\)  is therefore given by \(\mathbb {E}_{\mathcal {G}}[m]=pN\) . The weak law of large numbers implies that \(m\)  is tightly concentrated around its mean for large \(n\) . Therefore, we have \(m=\big ( p/2+o(1)\big )n^2\)  w.h.p. If \(A_n\)  denotes the adjaceny matrix of \(G_n\) , then \(\mathpzc {p}_2=\operatorname{tr}(A_n^2)=2m\)  [1]}, which implies that w.h.p.,
\(\mathpzc {p}_2=\big (p+o(1)\big )n^2.\) 
 If \(i>2\)  is even, then consider the bound \(|\lambda _1|^i\le \mathpzc {p}_i\le |\lambda _1|^i+n|\lambda _2|^i.\) 
Inequalities (REF ) and (REF ) imply that w.h.p.,
\(\big (p^i+o(1)\big )n^i\le \mathpzc {p}_i<\big (p^i+o(1)\big )n^i+O(n^{1+i/2}).\) 
 Observe that \(n^{1-i/2}\rightarrow 0\)  as \(n\rightarrow \infty \)  since \(i>2\) . We conclude that w.h.p.,
\(\mathpzc {p}_i=\big ( p^i+o(1)\big )n^i.\) 
 Let \(\pi =(j_1,j_2, \ldots , j_r)\)  be a partition of \(j\) . Lemma REF  together with relations (REF ) and (REF ) imply that w.h.p,
\(\mathpzc {p}_{j_1}\mathpzc {p}_{j_2}\cdots \mathpzc {p}_{j_r}=o(n^j)\) 
 whenever \(\pi \)  contains an odd integer larger than one. If \(m_1(\pi )\ne 0\) , then (REF ) still holds since \(\mathpzc {p}_1=\operatorname{tr}(A_n)=0\)  for simple graphs. Any partition of an odd integer must contain an odd part. Lemma REF  implies that w.h.p.,
\(f_j(\lambda )=o(n^j)\) 
 whenever \(j\)  is odd. This proves (a). If \(j\)  is even, then \(\mathpzc {p}_{j_1}\mathpzc {p}_{j_2}\cdots \mathpzc {p}_{j_r}=o(n^j)\)  w.h.p. unless \(\pi \)  is a partition of \(j\)  into even parts. Lemma REF , together with relations (REF ) and (REF ), imply that w.h.p.,
\(f_j(\lambda )=\Bigg ( j! \sum _{\pi \in E(d)} \frac{\kappa _{\pi }}{y_{\pi }} p^{j-m_2}+o(1) \Bigg )n^j,\) 
 which proves (b). We remark that the \(m_2\)  term appearing in the last expression occurs because of the discrepency in the power of \(p\)  occuring in (REF ) and (REF ).
